The function f(x) = 2x^3 - 9ax^2 + 12a^2x + 1 | vedclass

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(B) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. First,find the derivative $f'(x) = 6x^2 - 18ax + 12a^2$. Set $f'(x) = 0$ to find critical p

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$2$

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(B) Given the function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$. First,find the derivative $f'(x) = 6x^2 - 18ax + 12a^2$. Set $f'(x) = 0$ to find critical points: $6(x^2 - 3ax + 2a^2) = 0$. Factoring the quadratic gives $6(x - a)(x - 2a) = 0$,so the critical points are $x = a$ and $x = 2a$. Find the second derivative $f''(x) = 12x - 18a$. At $x = a$,$f''(a) = 12a - 18a = -6a  0$),so $x = a$ is a local maximum. Thus,$p = a$. At $x = 2a$,$f''(2a) = 12(2a) - 18a = 6a > 0$ (since $a > 0$),so $x = 2a$ is a local minimum. Thus,$q = 2a$. Given the condition $p^2 = q$,we substitute the values: $a^2 = 2a$. Since $a > 0$,we can divide by $a$ to get $a = 2$.

The function $f(x) = 2x^3 - 9ax^2 + 12a^2x + 1$ where $a > 0$ attains its local maximum and local minimum at $p$ and $q$ respectively. If $p^2 = q$,then $a =$

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